GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 12 Dec 2019, 16:32 ### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

#### Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.  # M21-02

Author Message
TAGS:

### Hide Tags

Math Expert V
Joined: 02 Sep 2009
Posts: 59712

### Show Tags

2
11 00:00

Difficulty:   85% (hard)

Question Stats: 53% (02:01) correct 47% (02:05) wrong based on 137 sessions

### HideShow timer Statistics

What is the median of set $$S =\{a - b, b - a, a + b\}$$ ?

(1) The mean of set $$S$$ is equal to $$a + b$$.

(2) The range of set $$S$$ is equal to $$2b$$.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59712

### Show Tags

1
2
Official Solution:

(1) The mean of set $$S$$ is equal to $$a + b$$. Given that $$mean=\frac{(a - b)+(b - a)+(a + b)}{3}=a+b$$, which leads to $$a+b=0$$. Now, if $$a+b=0$$, then $$a-b$$ and $$b-a$$ are either both zeros (if $$a=b=0$$) or have different signs (if $$a \ne b$$). In any case the median of $$S$$ is $$a+b=0$$. Sufficient.

(2) The range of set $$S$$ is equal to $$2b$$. If $$a=b=0$$, then $$median=0$$ but if $$a=0$$ and $$b=1$$, then $$median=1$$. Not sufficient.

_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59712

### Show Tags

1
1
Manmeet0 wrote:
Bunuel wrote:
manhasnoname wrote:
Option (1) states mean is a+b, which can be inferred from the question stem.

By simplifying this: $$mean=\frac{(a - b)+(b - a)+(a + b)}{3}=a+b$$

a+B/3=a+b ????????????

Yes, but we can simplify further:

(a + b)/3 = a + b;

a + b = 3(a + b);

2(a + b) = 0;

a + b = 0.
_________________
Math Expert V
Joined: 02 Sep 2009
Posts: 59712

### Show Tags

1
arven wrote:
I worked the second statement as following:

Range= Highest Number - Lowest Number
R=2b
In order for R to be 2b, the only possible combination is (b+a)-(a-b).
Thus, (a-b)≤(b-a)<(b+a) or (a-b)<(b-a)≤(b+a)
So b-a is the median.

Where am I wrong?

(2) The range of set $$S$$ is equal to $$2b$$. If $$a=b=0$$, then $$median=0$$ but if $$a=0$$ and $$b=1$$, then $$median=1$$. Not sufficient.
_________________
Intern  Joined: 20 Sep 2012
Posts: 3
GMAT 1: 670 Q48 V33 WE: Project Management (Consulting)

### Show Tags

In option 2, The range (largest - Least )= 2b; how would be arrive at 2b from the set of the three numbers viz : a+b, b-a, a-b ?
Math Expert V
Joined: 02 Sep 2009
Posts: 59712

### Show Tags

cjhande wrote:
In option 2, The range (largest - Least )= 2b; how would be arrive at 2b from the set of the three numbers viz : a+b, b-a, a-b ?

It's explained in the solution above:

If a = b = 0, then S = {0, 0, 0} --> the range = 0 - 0 = 0 = 2b.
If a = 0 and b = 1, then S = {-1, 1, 1} --> the range = 1 - (-1) = 2 = 2b.
_________________
Current Student Joined: 04 Jul 2016
Posts: 1
Location: India
GMAT 1: 750 Q51 V40 GPA: 3
WE: General Management (Venture Capital)

### Show Tags

a+b - (a-b) = 2b, which is true for a=b=0 as well. Hence that makes b-a as the median. From first option I got a+b as the median. What am I missing here?
Manager  S
Joined: 21 Apr 2016
Posts: 162

### Show Tags

Option (1) states mean is a+b, which can be inferred from the question stem.

Math Expert V
Joined: 02 Sep 2009
Posts: 59712

### Show Tags

manhasnoname wrote:
Option (1) states mean is a+b, which can be inferred from the question stem.

By simplifying this: $$mean=\frac{(a - b)+(b - a)+(a + b)}{3}=a+b$$
_________________
Intern  Joined: 10 Apr 2016
Posts: 1

### Show Tags

Bunuel wrote:
manhasnoname wrote:
Option (1) states mean is a+b, which can be inferred from the question stem.

By simplifying this: $$mean=\frac{(a - b)+(b - a)+(a + b)}{3}=a+b$$

a+B/3=a+b ????????????
Manager  B
Joined: 26 Mar 2016
Posts: 68
Location: Greece
GMAT 1: 710 Q51 V34 GPA: 2.9

### Show Tags

I worked the second statement as following:

Range= Highest Number - Lowest Number
R=2b
In order for R to be 2b, the only possible combination is (b+a)-(a-b).
Thus, (a-b)≤(b-a)<(b+a) or (a-b)<(b-a)≤(b+a)
So b-a is the median.

Where am I wrong?
Current Student B
Joined: 23 Nov 2016
Posts: 70
Location: United States (MN)
GMAT 1: 760 Q50 V42 GPA: 3.51

### Show Tags

Judging by this question, you can't assume that the set provided to you is in ascending order? I'd feel more comfortable removing that assumption if anyone can provide an OG question which has a set of unknown variables not provided in ascending order.
Math Expert V
Joined: 02 Sep 2009
Posts: 59712

### Show Tags

brooklyndude wrote:
Judging by this question, you can't assume that the set provided to you is in ascending order? I'd feel more comfortable removing that assumption if anyone can provide an OG question which has a set of unknown variables not provided in ascending order.

A set, by definition, is a collection of elements without any order. (While, a sequence, by definition, is an ordered list of terms.)
_________________
Manager  S
Joined: 27 Nov 2016
Posts: 53
Location: India
GPA: 2.71
WE: Consulting (Consulting)

### Show Tags

Did not understand the inference of 1st option.

How the median of the set is a+b ??

looking at the set median is b-a .
Intern  B
Joined: 24 Jun 2013
Posts: 2
Schools: CBS '17
GMAT 1: 630 Q47 V27 ### Show Tags

I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Hello expert, I get your explanation for option B, but aren't we worried about finding the mean in terms of a and b.

as explained even if a=b=0 or a=1, b=0, mean is b-a.
Math Expert V
Joined: 02 Sep 2009
Posts: 59712

### Show Tags

nitin083 wrote:
I think this is a high-quality question and the explanation isn't clear enough, please elaborate. Hello expert, I get your explanation for option B, but aren't we worried about finding the mean in terms of a and b.

as explained even if a=b=0 or a=1, b=0, mean is b-a.

The point is that in data sufficiency problems that ask for the value of a quantity, the data given in the statements are sufficient only when it is possible to determine exactly one numerical value for the quantity.
_________________
Senior Manager  S
Joined: 08 Jun 2015
Posts: 418
Location: India
GMAT 1: 640 Q48 V29 GMAT 2: 700 Q48 V38 GPA: 3.33

### Show Tags

+1 for A. My take :

Statement 1 : Given that mean is a+b. (a+b)/3=a+b ; i.e a+b=0. This means that the set is -2b,0,2b. The median in this case becomes zero. We get a definite answer.

Statement 2 : Range is 2b. The middle value is b-a. This is not a definite value. Hence not sufficient.

_________________
" The few , the fearless "
Manager  G
Joined: 15 Nov 2015
Posts: 158
Location: India
GMAT 1: 700 Q47 V39 GPA: 3.7

### Show Tags

Bunuel wrote:
Official Solution:

(1) The mean of set $$S$$ is equal to $$a + b$$. Given that $$mean=\frac{(a - b)+(b - a)+(a + b)}{3}=a+b$$, which leads to $$a+b=0$$. Now, if $$a+b=0$$, then $$a-b$$ and $$b-a$$ are either both zeros (if $$a=b=0$$) or have different signs (if $$a \ne b$$). In any case the median of $$S$$ is $$a+b=0$$. Sufficient.

(2) The range of set $$S$$ is equal to $$2b$$. If $$a=b=0$$, then $$median=0$$ but if $$a=0$$ and $$b=1$$, then $$median=1$$. Not sufficient.

chetan2u,, Bunuel

WRT option-A, if A+B=0, then how is median =0.

Do we assume that mean = median

kindly explain

Thanks
Manager  G
Joined: 15 Nov 2015
Posts: 158
Location: India
GMAT 1: 700 Q47 V39 GPA: 3.7

### Show Tags

Ashokshiva wrote:
Bunuel wrote:
Official Solution:

(1) The mean of set $$S$$ is equal to $$a + b$$. Given that $$mean=\frac{(a - b)+(b - a)+(a + b)}{3}=a+b$$, which leads to $$a+b=0$$. Now, if $$a+b=0$$, then $$a-b$$ and $$b-a$$ are either both zeros (if $$a=b=0$$) or have different signs (if $$a \ne b$$). In any case the median of $$S$$ is $$a+b=0$$. Sufficient.

(2) The range of set $$S$$ is equal to $$2b$$. If $$a=b=0$$, then $$median=0$$ but if $$a=0$$ and $$b=1$$, then $$median=1$$. Not sufficient.

chetan2u,, Bunuel

WRT option-A, if A+B=0, then how is median =0.

Do we assume that mean = median

kindly explain

Thanks

Got it

overlooked the given set.

Thanks M21-02   [#permalink] 08 Apr 2019, 08:35
Display posts from previous: Sort by

# M21-02

Moderators: chetan2u, Bunuel   